Finite differences and Factorials

by always_correct, Dec 24, 2016, 11:05 PM

Denote by $\Delta_{k}$ the $k$ th finite difference of a sequence. Show that $\Delta_{n} x^{n} = n!$ .

Clearly $\Delta_{1} x^{1} = 1!$ . Now we show
$\Delta_{n} x^{n} = n! \implies \Delta_{n+1} x^{n+1} = (n+1)!$
Assume that $\Delta_{n} x^{n} = n!$ , which holds for all $x$ . Take the anti-derivative of both sides. We get
$\begin{align*} \int \Delta_{n} x^{n} dx = \int n! \, dx \\ \Delta_{n} \int x^{n} dx = n!\cdot x + C_{1} \\ \Delta_{n} \frac{x^{n+1}}{n+1} + C_{2} = n!\cdot x + C_{1} \end{align*}$
Setting $x=0$ we see that $C_1 = C_2$ . We remove them from the equality.

$\begin{align*} \Delta_{n} \frac{x^{n+1}}{n+1} = n!\cdot x \\ \frac{1}{n+1} \cdot \Delta_{n} x^{n+1} = n! \cdot x \\ \Delta_{n} x ^{n+1} = (n+1)!\cdot x \end{align*}$
We then proceed to take the next finite difference.

$\begin{align*} \Delta_{n+1} x^{n+1} = \Delta_{1} (n+1)! \cdot x \\ \Delta_{n+1} x^{n+1} = (n+1)! \cdot (x+1) - (n+1)! \cdot x \end{align*}$ $\boxed{\Delta_{n+1} x^{n+1} = (n+1)!}$

This post has been edited 2 times. Last edited by always_correct, Dec 24, 2016, 11:06 PM

Finite Differences

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Finite differences and Factorials

by always_correct, Dec 24, 2016, 11:05 PM

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